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4 years ago

Help plzzzz! pre calc question

Mathematics

1 answer:

Mumz [18]4 years ago

7 0

Answer:

$\sum^{\infty}_0 {(-3)^n}$

Step-by-step explanation:

Notice that the series: 1 - 3 + 9 - 27 +... clearly has powers of the factor 3 in its terms and it is also an alternate series (the terms alternate from positive to negative). The terms are positive for $a_0, a_2, a_4, ...$ (even terms) , while the odd terms $a_1, a_3, a_5, ...$ are negative. (so most likely there should be a factor (-1) in the common ratio.

We can then represent it with the following summation expression:

$\sum^{\infty}_0 {(-3)^n}$ given that each of its first four terms are:

$a_0=(-3)^0=1\\a_1=(-3)^1=-3\\a_2=(-3)^2=(-3)*(-3)=9\\a_3=(-3)^3=(-3)*(-3)*(-3)=-27$

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Which expression is equivalent to 10p - 35?

s344n2d4d5 [400]

Answer:

a: $5\left(2p-7\right)$

Step-by-step explanation:

$10p-35$

$5* \:2p-5* \:7$

$5\left(2p-7\right)$

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3 years ago

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Find all real solutions to the equation (x² − 6x +3)(2x² − 4x − 7) = 0.

Jet001 [13]

Answer:

x = 3 + √6 ; x = 3 - √6 ; $x = \frac{2+3\sqrt{2}}{2}$ ; $x = \frac{2-(3)\sqrt{2}}{2}$

Step-by-step explanation:

Relation given in the question:

(x² − 6x +3)(2x² − 4x − 7) = 0

Now,

for the above relation to be true the following condition must be followed:

Either (x² − 6x +3) = 0 ............(1)

(2x² − 4x − 7) = 0 ..........(2)

now considering the equation (1)

(x² − 6x +3) = 0

the roots can be found out as:

$x = \frac{-b\pm\sqrt{b^2-4ac}}{2a}$

for the equation ax² + bx + c = 0

thus,

the roots are

$x = \frac{-(-6)\pm\sqrt{(-6)^2-4\times1\times(3)}}{2\times(1)}$

$x = \frac{6\pm\sqrt{36-12}}{2}$

$x = \frac{6+\sqrt{24}}{2}$ and, x = $x = \frac{6-\sqrt{24}}{2}$

$x = \frac{6+2\sqrt{6}}{2}$ and, x = $x = \frac{6-2\sqrt{6}}{2}$

x = 3 + √6 and x = 3 - √6

similarly for (2x² − 4x − 7) = 0.

we have

the roots are

$x = \frac{-(-4)\pm\sqrt{(-4)^2-4\times2\times(-7)}}{2\times(2)}$

$x = \frac{4\pm\sqrt{16+56}}{4}$

$x = \frac{4+\sqrt{72}}{4}$ and, x = $x = \frac{4-\sqrt{72}}{4}$

$x = \frac{4+\sqrt{2^2\times3^2\times2}}{2}$ and, x = $x = \frac{4-\sqrt{2^2\times3^2\times2}}{4}$

$x = \frac{4+(2\times3)\sqrt{2}}{2}$ and, x = $x = \frac{4-(2\times3)\sqrt{2}}{4}$

$x = \frac{2+3\sqrt{2}}{2}$ and, $x = \frac{2-(3)\sqrt{2}}{2}$

Hence, the possible roots are

x = 3 + √6 ; x = 3 - √6 ; $x = \frac{2+3\sqrt{2}}{2}$ ; $x = \frac{2-(3)\sqrt{2}}{2}$

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3 years ago

What two simpler problem can you use to find 9×38?

elena55 [62]

Just a shot in the dark: 9*30 and 9*8?

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3 years ago

Write the quadratic equation y=x^2-6+7 in vertex form

antoniya [11.8K]

Half the x-coefficient is -3, so the equation needs to be rearranged into a form that looks like
x² -6x +(-3)² + [something]
You can get there by adding and subtracting 9 from the original equation.
y = x² -6x +9 +7 -9 . . . . . 9 added and subtracted
y = (x² -6x +9) -2
The quantity in parentheses is a perfect square, so we can write the equation in the desired form as ...
y = (x -3)² -2

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3 years ago

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AnnyKZ [126]

Answer:

what's the question ndndn

6 0

3 years ago

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